Question

A 2L container has 1 L water and 1 L air headspace. The
headspace has a total pressure of 1 atm. The temperature of both
water and air is 298K. 1 mg of formaldehyde (HCHO) was added into
the water. Since formaldehyde is a volatile compound, some
formaldehyde will escape into the air. The Henry’s constant (KH)
for formaldehyde is 6300 M?atm-1. What is the mass concentration of
formaldehyde in the water after the equilibrium between the air and
the water has been reached? What is the molar fraction of
formaldehyde (YHCHO) in the air headspace at equilibrium?

Answer #1

yP=xKH

y/x = 6300

where

y is mole fraction of formaldehyde in air haedspace

x is mole fraction of formaldehyde in water

M=molar mass of formaldehyde =30

m=mass of formaldehyde =1 mg

**n=moles of formaldehyde =0.033 mmol**

y/x = 6300

6301X =0.033

**X=5.2372*10 ^{-6} formaldehydemoles in liquid
phase**

**6300X =0.03299436 formaldehyde moles in gas
phase**

moles of water in 1 L

density of water is 1 kg/L

**so moles of water in 1 L =
1/18 = 0.055 kmol**

moles of air in 1 L space at 1 atm 298K

PV =nRT

101.325 *0.001 =n *8.314 *298

**n=4.08*10 ^{-5} kmol
=0.0408 mole of air**

so

mole fraction of formaldehyde in water =
**5.2372*10 ^{-6}/(0.055*1000+5.2372*10^{-6})
=9.52218 *10^{-8}**

mole fraction of formaldehyde in air = **0.03299436
/(0.0408+0.03299436 ) =0.4471**

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